Solving by factorisation

The quick­est way to solve a qua­dratic equa­tion, when it works, is to fac­torise the left side into two brack­ets. The method rests on one sim­ple fact about mul­ti­pli­ca­tion: if a prod­uct is zero, one of the fac­tors must be zero. This les­son uses the split­ting-the-mid­dle-term method from the chap­ter on fac­tori­sa­tion to solve qua­dratic equa­tions.

The zero prod­uct rule#

If $p \times q = 0$, then $p = 0$ or $q = 0$ (or both).

No two non-zero num­bers mul­ti­ply to give 0. So once a qua­dratic is writ­ten as $(x - \alpha)(x - \beta) = 0$, its roots can be read off: $x = \alpha$ or $x = \beta$.

This only works when one side is zero. From $(x - 2)(x - 3) = 6$ you can­not con­clude that $x - 2 = 6$ or any­thing sim­i­lar. Always bring the equa­tion to the form "expres­sion $= 0$" first.

Split­ting the mid­dle term#

To fac­torise $ax^2 + bx + c$, find two num­bers whose prod­uct is $ac$ and whose sum is $b$. Use them to split $bx$ into two terms, then group in pairs and take out com­mon fac­tors.

Exam­ple 1#

Solve $x^2 - 7x + 12 = 0$.

We need two num­bers with prod­uct $1 \times 12 = 12$ and sum $-7$: they are $-3$ and $-4$.

$$\begin{aligned} x^2 - 3x - 4x + 12 &= 0 \\ x(x - 3) - 4(x - 3) &= 0 \\ (x - 3)(x - 4) &= 0 \end{aligned}$$

So $x - 3 = 0$ or $x - 4 = 0$, giv­ing $x = 3$ or $x = 4$.

Check: $9 - 21 + 12 = 0$ and $16 - 28 + 12 = 0$.

Exam­ple 2#

Solve $2x^2 + 5x - 3 = 0$.

Prod­uct $ac = 2 \times (-3) = -6$, sum $5$: the num­bers are $6$ and $-1$.

$$\begin{aligned} 2x^2 + 6x - x - 3 &= 0 \\ 2x(x + 3) - 1(x + 3) &= 0 \\ (2x - 1)(x + 3) &= 0 \end{aligned}$$

So $2x - 1 = 0$ or $x + 3 = 0$, giv­ing $\displaystyle x = \frac{1}{2}$ or $x = -3$.

Exam­ple 3#

Solve $6x^2 - 13x + 6 = 0$.

Prod­uct $ac = 36$, sum $-13$: the num­bers are $-4$ and $-9$.

$$\begin{aligned} 6x^2 - 4x - 9x + 6 &= 0 \\ 2x(3x - 2) - 3(3x - 2) &= 0 \\ (2x - 3)(3x - 2) &= 0 \end{aligned}$$

So $\displaystyle x = \frac{3}{2}$ or $\displaystyle x = \frac{2}{3}$. Notice that the two roots are rec­i­p­ro­cals of each other; this hap­pens because $a = c$.

Exam­ple 4#

Solve $x^2 + 3x - 108 = 0$ (the rec­tan­gle from the last les­son).

Prod­uct $-108$, sum $3$: the num­bers are $12$ and $-9$.

$$x^2 + 12x - 9x - 108 = 0 \implies x(x + 12) - 9(x + 12) = 0 \implies (x - 9)(x + 12) = 0$$

So $x = 9$ or $x = -12$. A width can­not be neg­a­tive, so the width is 9 m and the length is 12 m.

Two spe­cial shapes#

No con­stant term. When $c = 0$, take out the com­mon fac­tor $x$.

Exam­ple 5#

Solve $x^2 = 5x$.

Bring every­thing to one side: $x^2 - 5x = 0$, so $x(x - 5) = 0$, giv­ing $x = 0$ or $x = 5$.

Do not divide both sides by $x$. That would give only $x = 5$ and lose the root $x = 0$. Divid­ing by an expres­sion that might be zero throws away solu­tions.

No mid­dle term. When $b = 0$ and the con­stant is neg­a­tive, use the dif­fer­ence of two squares, $p^2 - q^2 = (p - q)(p + q)$.

Exam­ple 6#

Solve $9x^2 - 25 = 0$.

$$\displaystyle (3x)^2 - 5^2 = 0 \implies (3x - 5)(3x + 5) = 0 \implies x = \frac{5}{3} \text{ or } x = -\frac{5}{3}$$

A quick alter­na­tive: $9x^2 = 25$, so $\displaystyle x^2 = \frac{25}{9}$ and $\displaystyle x = \pm\frac{5}{3}$. Remem­ber both signs.

Equa­tions with surds#

Exam­ple 7#

Solve $x^2 - (\sqrt{3} + 1)x + \sqrt{3} = 0$.

We need two num­bers with prod­uct $\sqrt{3}$ and sum $-(\sqrt{3} + 1)$: they are $-\sqrt{3}$ and $-1$.

$$x^2 - \sqrt{3}x - x + \sqrt{3} = 0 \implies x(x - \sqrt{3}) - 1(x - \sqrt{3}) = 0 \implies (x - 1)(x - \sqrt{3}) = 0$$

So $x = 1$ or $x = \sqrt{3}$.

Equa­tions that need tidy­ing first#

Exam­ple 8#

Solve $\displaystyle \frac{x}{3} + \frac{2}{x} = \frac{5}{3}$, where $x \neq 0$.

Mul­ti­ply every term by $3x$: $x^2 + 6 = 5x$, that is $x^2 - 5x + 6 = 0$. This fac­torises as $(x - 2)(x - 3) = 0$, so $x = 2$ or $x = 3$. Nei­ther makes a denom­i­na­tor zero, so both are roots.

Exam­ple 9#

Solve $(x - 1)(x + 2) = 18$.

Expand and bring 18 across: $x^2 + x - 2 - 18 = 0$, so $x^2 + x - 20 = 0$. Prod­uct $-20$, sum $1$: the num­bers are $5$ and $-4$. So $(x + 5)(x - 4) = 0$, giv­ing $x = -5$ or $x = 4$.

Check $x = 4$: $3 \times 6 = 18$. Check $x = -5$: $(-6) \times (-3) = 18$.

Prac­tice#

Solve each equa­tion by fac­tori­sa­tion.

1. $x^2 + 9x + 20 = 0$
2. $x^2 - 2x - 35 = 0$
3. $3x^2 - 10x + 8 = 0$
4. $2x^2 - x - 15 = 0$
5. $4x^2 - 12x + 9 = 0$
6. $5x^2 = 20x$
7. $16x^2 - 49 = 0$
8. $x^2 - (\sqrt{2} + 2)x + 2\sqrt{2} = 0$
9. $\displaystyle x + \frac{12}{x} = 7$, where $x \neq 0$
10. $(2x + 1)(x - 3) = 15$

Answers#